码迷,mamicode.com
首页 > 其他好文 > 详细

【模板】多标记 LCT

时间:2019-09-21 23:21:13      阅读:110      评论:0      收藏:0      [点我收藏+]

标签:lin   its   type   lse   turn   static   tor   ++   delete   

代码如下

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

const int mod = 51061;

struct node {
  node *l, *r, *p;
  int rev, val;
  LL sum, add, mul, sz;
  node() {
    l = r = p = NULL;
    sum = add = rev = 0;
    mul = val = sz = 1;
  }
  void unsafe_reverse() {
    swap(l, r);
    rev ^= 1;
  }
  void unsafe_add(LL x) {
    sum = (sum + x * sz) % mod;
    add = (add + x) % mod;
    val = (val + x) % mod;
  }
  void unsafe_mul(LL x) {
    sum = sum * x % mod;
    add = add * x % mod;
    val = val * x % mod;
    mul = mul * x % mod;
  }
  void pull() {
    sum = val;
    sz = 1;
    if (l != NULL) {
      l->p = this;
      sum = (sum + l->sum) % mod;
      sz += l->sz;
    }
    if (r != NULL) {
      r->p = this;
      sum = (sum + r->sum) % mod;
      sz += r->sz;
    }
  }
  void push() {
    if (rev) {
      if (l != NULL) {
        l->unsafe_reverse();
      }
      if (r != NULL) {
        r->unsafe_reverse();
      }
      rev = 0;
    }
    if (mul != 1) {
      if (l != NULL) {
        l->unsafe_mul(mul);
      }
      if (r != NULL) {
        r->unsafe_mul(mul);
      }
      mul = 1;
    }
    if (add != 0) {
      if (l != NULL) {
        l->unsafe_add(add);
      }
      if (r != NULL) {
        r->unsafe_add(add);
      }
      add = 0;
    }
  }
};
bool is_root(node *v) {
  if (v == NULL) {
    return false;
  }
  return (v->p == NULL) || (v->p->l != v && v->p->r != v);
}
void rotate(node *v) {
  node *u = v->p;
  assert(u != NULL);
  v->p = u->p;
  if (v->p != NULL) {
    if (v->p->l == u) {
      v->p->l = v;
    }
    if (v->p->r == u) {
      v->p->r = v;
    }
  }
  if (v == u->l) {
    u->l = v->r;
    v->r = u;
  }
  if (v == u->r) {
    u->r = v->l;
    v->l = u;
  }
  u->pull();
  v->pull();
}
void deal_with_push(node *v) {
  static stack<node*> s;
  while (1) {
    s.push(v);
    if (is_root(v)) {
      break;
    }
    v = v->p;
  }
  while (!s.empty()) {
    s.top()->push();
    s.pop();
  }
}
void splay(node *v) {
  deal_with_push(v);
  while (!is_root(v)) {
    node *u = v->p;
    if (!is_root(u)) {
      if ((v == u->l) ^ (u == u->p->l)) {
        rotate(v);
      } else {
        rotate(u);
      }
    }
    rotate(v);
  }
}
void access(node *v) {
  node *u = NULL;
  while (v != NULL) {
    splay(v);
    v->r = u;
    v->pull();
    u = v;
    v = v->p;
  }
}
void make_root(node *v) {
  access(v);
  splay(v);
  v->unsafe_reverse();
}
node* find_root(node *v) {
  access(v);
  splay(v);
  while (v->l != NULL) {
    v->push();
    v = v->l;
  }
  splay(v);
  return v;
}
void link(node *v, node *u) {
  if (find_root(v) != find_root(u)) {
    make_root(v);
    v->p = u;
  }
}
void cut(node *v, node *u) {
  make_root(v);
  if (find_root(u) == v && u->p == v && u->l == NULL) {
    u->p = v->r = NULL;
    v->pull();
  }
}
void split(node *v, node *u) {
  make_root(v);
  access(u);
  splay(u);
}

int main() {
  //freopen("data.in", "r", stdin);
  ios::sync_with_stdio(false);
  cin.tie(0), cout.tie(0);
  int n, m;
  cin >> n >> m;
  vector<node*> t(n + 1);
  for (int i = 1; i <= n; i++) {
    t[i] = new node();
  }
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    link(t[x], t[y]);
  }
  while (m--) {
    string opt;
    int x, y;
    cin >> opt >> x >> y;
    if (opt[0] == '+') {
      int c;
      cin >> c;
      split(t[x], t[y]);
      t[y]->unsafe_add(c);
    }
    if (opt[0] == '-') {
      int u, v;
      cin >> u >> v;
      cut(t[x], t[y]);
      link(t[u], t[v]);
    }
    if (opt[0] == '/') {
      split(t[x], t[y]);
      cout << t[y]->sum % mod << endl;
    }
    if (opt[0] == '*') {
      int c;
      cin >> c;
      split(t[x], t[y]);
      t[y]->unsafe_mul(c);
    }
  }
  for (int i = 1; i <= n; i++) {
    delete t[i];
  }
  return 0;
}

【模板】多标记 LCT

标签:lin   its   type   lse   turn   static   tor   ++   delete   

原文地址:https://www.cnblogs.com/wzj-xhjbk/p/11565125.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!